If the equation $x^{4} + 4 x^{3} - 2 x^{2} - 12 x + 9 = 0$ has two pairs of equal roots, find the roots of the equation.

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Solution

Let the roots be $a, a, b, b$
we have $a + a + b + b = - 4$ and $a^{2} b^{2} = 9$
$\Rightarrow a + b = - 2$ and $a b = \pm 3$
Case I: $a b = 3$
$(a - b)^{2} = (a + b)^{2} - 4 a b$
$(a - b)^{2} = 4 - 12 = - 8$ which is not possible. Hence $a b \neq 3$
Case II: $a b = - 3$
$(a - b)^{2} = (a + b)^{2} - 4 a b$
$(a - b)^{2} = 4 - 4 (- 3) = 4 + 12 = 16$
$(a - b) = \pm 4$

$a + b = - 2$
$a - b = 4$ or $a - b = - 4$
$\Rightarrow 2 a = 2$ or $2 a = - 6$
$\Rightarrow a = 1$ or $a = - 3$
$\Rightarrow b = - 3$ or $b = 1$
Hence we get $- 3, - 3, 1, 1$ as roots of given equation

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